On my honor as a student, I have neither given nor received unauthorized assistance on this exam.

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1 HW2 (Overview of Transport) (Print name above) On my onor as a student, I ave neiter given nor received unautorized assistance on tis exam. (sign name above) 1

2 Figure 1: Band-diagram before and after application of V SG > 0 1. Problem E1.2 of te book (Transistor I-Vs). (a) Plot te I d V d caracteristics for a silicon MOSFET wit parameters given in te book. Plot it for Vg=0.25V and Vg=0.5V. You can use te code at te back of te book as a guide. Mark canges on te code for te following parts. [7] Te band-diagram sows wat s appening wit V g. Positive gate bias lowers levels relative to te contacts, until te empty conduction band reaces te Fermi energy. At tat voltage, current starts flowing under a small V SD. Tis explains te tresold beavior on te left. For te rigt figure, we start initial at low V SD wit bot electrocemical potentials inside te contacts in teir valence bands. For positive drain bias, current starts flowing troug tis band near te Fermi energy, and te current keeps increasing as te bias window expands over te density of states, until te drain electrocemical potential µ D goes down all te way into te band-gap. At tis point, tere are no new states tat enter te window, and no more current can be drawn wit increasing V SD, wereupon te current saturates. (b) Wat appens to te current wen you cange te gate control parameter α g? Alter te ratio C S /C G = C D /C G from 0.05 to 0.01 and ten to 0.1 and see if you can rationalize tis beavior. [5] Te cannel slips because of te local Laplace potential U L = qv, were V is te capacitively weigted average of te various contact potentials. Normally C G C D and we ave te above saturating case were te drain potential simply slips past a non-slipping cannel density of states. Wen C D increases, te gate loses control and te cannel starts slipping, because te fraction of V tat varies wit V SD increases. Te drain must now over- 2

3 take a moving band-edge in order to saturate, and since te band continues to slip, tis draws out te saturation process into a non-saturating current (c) Wat appens to te slope of te I-Vg if you increase te ratio C S /C G and wy? Describe two ways in wic you can decrease tis ratio. [3+2] Te I V g also canges, because te current canges under V SD even wen V SG is fixed, especially wen C S is large. Te slippage makes it easier for te level to reac te Fermi energy, and te tresold canges wit drain bias. Te slope of te I V g (related to te subtresold swing, te inverse slope), also canges, because it takes a larger amount of gate voltage to turn te cannel on for a fixed drain bias, because part of te potential is being siponed into te drain control. Tis makes te I V g less steep. We can cange te C D /C G ratio by eiter playing wit te oxide tickness relative to te cannel lengt, or te oxide composition (specifically, dielectric constant ɛ) relative to te cannel, since a parallel plate capacitor as a capacitance per unit area given by dielectric constant divided by plate separation, from our EM lectures. (d) Is tis an n-type or p-type transistor? [3] n-type, because bringing te levels down wit a positive gate bias populates formerly empty levels wit electrons 3

4 Figure 2: A eat engine 2. Termocurrent A cannel wit a single sarp level ɛ 0 is connected to a ot contact to te left at temperature T 1 and a cold contact to te rigt at temperature T 2. No voltage is applied so tat we are under open circuit conditions. In oter words, we start wit µ 1 = µ 2 = E F, but since te temperatures are unequal, f 1 f 2. (a) Wic way does te electron current flow, assuming ɛ 0 > E F? [5] More electrons on otter contact for iger energy, so electrons flow from otter to colder contact (b) As electrons flow and build up at one of te contacts, te electron number at tat contact increases as tere is no battery to sipon te carges away. Tus te electrocemical potential µ at tat contact increases relative to te oter contact, and a net open circuit voltage builds up. Te maximum open circuit voltage V oc is reaced wen te current stops flowing, in oter words, wen f 1 = f 2 (See picture below). Write down te equation connecting µ 1 and µ 2 at tat point. [10] f 1 = f 2 means (ɛ 0 µ 1 )/T 1 = (ɛ 0 µ 2 )/T 2 (c) Te eat extracted from te ot contact is Q 1 = ɛ 0 µ 1, wile te 4

5 eat dumped into te cold contact is Q 2 = ɛ 0 µ 2. Tus, te work done W = qv oc is related to Q 1 and Q 2. Sow tat te maximum efficiency (output to input energy ratio) of te conversion of eat to work is given by te Carnot limit, W/Q 1 = 1 T 2 /T 1.[10] Te above equation says Q 1 /T 1 = Q 2 /T 2 (tis is actually an ideal equal entropy situation). Also, W = Q 1 Q 2 from energy conservation, so we get W/Q 1 = 1 Q 2 /Q 1 = 1 T 2 /T 1. Tis actually is already te best case. Tis is te Carnot efficiency. We assume te system as reaced equilibrium wen we calculate tis efficiency (since we set f 1 = f 2 ). It also ignores any eat flux tat dissipates into te system troug ponon conduction. Tis means tat we assume tat te entire eat generated is at te junctions, driven by electron currents alone (ie, Peltier eat), wile ponon conduction of eat (ie, energy carried not by electrons but by atomic motion) is zero. Te ratio of te Peltier to te ponon conducted eat is te termoelectric figure of merit, ZT. We assume tat to be infinite ere, wic gives us te Carnot limit. Finite ZT will bring tat efficiency down. Te next problem deals wit tis! 5

6 Pb3. Seebeck coefficient and termoelectric figure of merit Let s go back to our basic equation I = 2q dem(e)[f 1 (E) f 2 (E)] (1) were M(E) is te mode spectrum. Te quantity f 1 f 2 is non-zero and is te driver of current. It drives current because of a voltage difference V and a temperature difference T across te junction. We can tus linearize tis equation as I = G V + G S T (2) were G and G S can be extracted from te integral (for instance, it s easy to sow by Taylor expanding tat G = de( f/ E)[2q 2 M(E)/]). But our aim is to go a bit furter. (a) Te sort circuit current I SC is obtained by setting te voltage V to zero. Te open circuit voltage V OC is obtained by setting te curent I to zero. Find te Seebeck coefficient, defined as te ratio S = V OC / T, in terms of te extracted parameters G (te conductance ) and G S. I SC = G S T. V OC = G S T/G Tus, S = V OC / T = G S /G 6

7 An aside on Seebeck and Peltier It is easy to do Taylor expansion and sow tat G = I V = 2q dem(e) f ( ) E E F k B T V = 2q2 dem(e) f E (3) using sift in E by -qv. Similarly, G S = I T = 2q = 2q dem(e) f ( ) E E F k B T T ( ) dem(e) f E E F x k B T 2 = 2q dem(e) ( E E F T ) f E (4) 7

8 We could actually do te same for te power, i.e., energy current I Q = 2 de(e E F )M(E)[f 1 (E) f 2 (E)] Te two Taylor expansions are I Q = 2 de(e E F )M(E) f V V = 2q de(e E F )M(E) f E = G S T (5) so tat I Q = ( G S T ) V + κ T (6) Note tat κ is not te termal conductivity, wic is not supposed to be te sort circuit ( V = 0) result but te open circuit (I = 0) result. So we need to rewrite it in terms of I and T, to get ( ) I Q = G S T/G } {{ } Π ( I + ) κ + G 2 ST/G T } {{ } G K Π is te Peltier coefficient, and G K is te termal conduction (b) Let us connect tis open circuit device to drive an external load. Te best power transfer to te load will appen wen te load resistance equals te cannel resistance (R = 1/G). Under tat condition wit bot matced resistors, te maximum power transferred can be easily sown from elementary circuit teory to be P max = V 2 OC/4R = GV 2 OC/4. (7) Te input power is te eat current, defined as P in = G K T (8) were G K is te termal conductance (just like carge current is carge per unit time, eat current is eat energy per unit time, and energy/time is power). 8

9 Sow tat te termoelectric efficiency can be written as ( P max S 2 G = T )( ) T P in G K 4 T (9) were T = (T 1 + T 2 )/2 is te average contact temperature. Te first factor on te rigt is called te termoelectric figure of merit, Z T V OC = S T by definition of S, so tat P max /P in = G( S T ) 2 /4G K T = S 2 G T/4G K, wic can be cast in te form given by multiplying numerator and denominator by T, ie, P max /P in = (Z T )( T/4 T ) Note tat We sould ave substracted power back-delivered to te load at te ot contact in te denominator ( I 2 R load /2) and also te Peltier eat ΠI sown above, generated by current flow alone and not by ponon conduction G K. Also, we assumed impedance matcing for maximum power transfer, wic is different from maximum efficiency (since max power transfer to load also increases power backflow to contacts). Once we do tese two, te efficiency canges to P max P in = ( T T 1 ) }{{} Carnot ( ) 1 + Z T Z T + T2 /T 1 (10) For small Z T, we can Taylor expand, giving us on te rigt side te expression I asked you to derive in Eq. 5. Note tat we reac Carnot efficiency wen te termoelectric figure of merit ZT 1. Termoelectricity is expected to be a viable source of energy conversion if te number ZT > 3. Tis will require making G large and G K small. Since G (carge conductance) is usually due to electrons and G K (eat conductance) in insulators is usually due to vibrations (ponons), te trick is to make crystals tat are conductive to electrons and insulating to ponons (te so-called electron-crystal, ponon-glass) 9

10 Figure 3: Contact due to scattering Problem 4 Incoerent scattering (grad students only) Te lectures in class assumed tat an electron runs troug te cannel in one step, also referred to as coerent scattering processes. Let us see if we can now include incoerent scattering in our model (were te electron pauses inside te cannel before exiting). We refer to Fig. 2. As before, we ave a density of states D(E) and two contacts wit injection rates γ 1,2 and electrocemical potentials µ 1,2 tat define teir Fermi functions f 1,2 (E).We will treat te scattering center sas a tird virtual contact wit its own Fermi function f s (E) and injection rate γ s. We can write tree transmission functions, one between contacts 1 and 2 given by T 12 (E) = 2π[γ 1 γ 2 /(γ 1 + γ 2 )]D(E), one between contacts 1 and s given by T 1S (E) = 2π[γ 1 γ S /(γ 1 + γ S )]D(E), and one between contacts 2 and S given by T 2S (E) = 2π[γ 2 γ S /(γ 2 + γ S )]D(E). We can ten calculate current at te m t terminal (m = 1, 2 or s), by summing all outgoing currents troug it (tis is a generalization of te two-terminal Landauer formula, called te Landauer-Büttiker formula). 10

11 I m = (2q/) de[t m1 (f m f 1 ) + T m2 (f m f 2 ) +...] (11) = (2q/) det mn (f m f n ) (12) mn Let us assume we know γ s (determined by some microscopic process), but we still do not know f s (E). Te contact s acts like a regular contact, except it does not draw any net current (as tere is no real terminal tere). Any electron witdrawn at te contact is reinjected but in te process it loses memory (coerence). For instance, if an electron enters contact s from 1, it is promptly reinjected back into te cannel, but now te injected electron can go eiter way (from contact s back to 1 or onwards to 2). In oter words, just by adding a virtual contact at s we ave made te electron pause, and lose its memory of were it was going. (a) Impose te condition tat te contact s does not draw any net current, to derive a relation between f 1, f 2 and f s, using te equations above. You will get an equation involving energy integrals. [6] [ ] I S = 0 means de T 1S (f 1 f s ) + T 2S (f 2 f S ) = 0 (b) Let us assume tat altoug te electron loses its memory, it still does not lose energy (in oter words, te scattering center is unable to add to or remove any energy from te reinjected electron). Tis is called elastic scattering. Tus, energy levels between electrons do not mix. Tis means tat you sould be able to pull te integrands from te above equation out of te integral. Write down an equation for f s (E) in terms of te oter parameters given to you. Solve for f s (E) and simplify [6] f S = T 1Sf 1 + T 2S f 2 T 1S + T 2S Every term ere is energy dependent, wit te same value of te energy argument E. Most importantly, for large enoug drain bias V SD k B T, it s easy to sow tat tis function f S does not ave te form of a Fermi- 11

12 Figure 4: f S for T 1S Fermi-Dirac. = T 2S sows nonequilibrium jumps deviating from Dirac distribution, but as multiple jumps. It is patently non-equilibrium, because coming back to equilibrium would require rearranging te energy cannels tat we re excluding troug our ignorance of energy-dependent scattering (c) Now let us calculate te terminal current I 1 (E) using te above summed equation, substituting te expression for f s (E) tat you just derived. Sow tat te current now as te same form as a two-terminal current, in oter words, I 1 = (2q/) de T (f 1 f 2 ) Write down a compact expression for T. [8] I 1 = (2q/) [ ] de T 1S (f 1 f S ) + T 12 (f 1 f 2 ) Substituting f S from above, we get wit some simple algebra, I 1 = (2q/) de [ T 12 + T 1ST 2S T 1S + T }{{ 2S } T ] (f 1 f 2 ) (13) Tis can be written as te circuit model above. Note tat we aven t included any energy-dependent scattering ere. We 12

13 Figure 5: Equivalent circuit model - a ballistic cannel below, in parallel wit a scattering cannel tat is a series combination of a ballistic pat to te scatterer and a ballistic pat tereafter. Eac resistor is te inverse of its transmission, as expected! can do tat if we do not simplify our equation by pulling te integrand out of te integral as we did, but try to solve te equation in a different way by making an assumption about te nature of f s. For instance, if we assume tat te contact actually randomizes te electron energies and tis energy mixing actually brings te electrons into local equilibrium, ten we can postulate an actual form for f S wit is a Fermi-Dirac function (unlike te previous part). Ten we just ensure te integral I S = 0 by varying µ S till tat appens, and ten use tat f S tereafter (tis can only be done numerically) 13